# Power Series&Function Series Questions

• Jan 18th 2010, 11:56 PM
WannaBe
Power Series&Function Series Questions
I'll be delighted to get Some help in these questions:

1. Determine whether the next function series uniformly converges:
SIGMA_ (1-x)^2 * x^n in [0,1]

2. Determine where the next series converges and what is the sum of the series: SIGMA_(nx-n+1)*x^n

Thanks a lot!
• Jan 19th 2010, 12:19 AM
CaptainBlack
Quote:

Originally Posted by WannaBe
I'll be delighted to get Some help in these questions:

1. Determine whether the next function series uniformly converges:
SIGMA_ (1-x)^2 * x^n in [0,1]

First you should be able to recognise what the sum of this series is:

$\displaystyle \sum_{n=0}^{\infty} (1-x)^2 x^n= (1-x)$

So now we are interested in:

$\displaystyle A(N,x)=\left| \left(\sum_{n=0}^N (1-x)^2 x^n\right) - (1-x) \right|$

for $\displaystyle x \in (0,1)$.

Simplify:

$\displaystyle A(N,x)=(1-x)x^{N+1}$

Now you need to determine if the supremum of $\displaystyle A(N,x)$ on $\displaystyle (0,1)$ is bounded by something independednt of $\displaystyle x$ that goes to zero as $\displaystyle N \to \infty$

To do this find the maximum of $\displaystyle A(N,x)$ for $\displaystyle x \in [0,1]$ and show that this goes to zero as $\displaystyle N$ goes to infinity.

CB
• Jan 19th 2010, 12:23 AM
CaptainBlack
Quote:

Originally Posted by WannaBe
2. Determine where the next series converges and what is the sum of the series: SIGMA_(nx-n+1)*x^n

Start by finding the sum of the series.

CB
• Jan 19th 2010, 03:44 AM
WannaBe
Hey Captain Black,
Thanks for your fast answer... In the first question we were able to determine the sum of the series by the formula for infinite geometric series sum that its "ratio" is less then 1...
How can we do the same thing for the second series?
THanks
• Jan 19th 2010, 04:10 AM
CaptainBlack
Quote:

Originally Posted by WannaBe
Hey Captain Black,
Thanks for your fast answer... In the first question we were able to determine the sum of the series by the formula for infinite geometric series sum that its "ratio" is less then 1...
How can we do the same thing for the second series?
THanks

You need to find:

$\displaystyle S_1=\sum_{n=0}^{\infty} n x^n$

then the sum can be writen in terms of this and:

$\displaystyle S_2=\sum_{n=0}^{\infty} x^n$.

The second of these we know. The first we can find by considering:

$\displaystyle D=\frac{d}{dx}\sum_{n=1}^{\infty} x^n$

CB
• Jan 19th 2010, 07:06 AM
WannaBe
Let's see:
$\displaystyle S_1=\sum_{n=1}^{\infty} n x^n$ converges for every x in (-1,1) and diverges for every x such as |x|>=1.

$\displaystyle S_2=\sum_{n=1}^{\infty} x^n$converges for ever x in (-1,1) and diverges for every x such as: |x|>=1.

What should I do next? We have another element in the sum :
$\displaystyle S_3 = \sum_{n=1}^{\infty} n \times x^{n+1}$ and I've no idea what to do with it...

Thanks!
• Jan 19th 2010, 07:23 AM
WannaBe
The series actually goes from 1 to infinity...
Hence:
$\displaystyle \sum_{n=1}^{\infty} (1-x)^2 x^n = \frac{x}{1-x}$
And:
$\displaystyle A(N,x)=\left| \left(\sum_{n=0}^N (1-x)^2 x^n\right) - \frac{x}{1-x} \right|$

If we simplify a little bit, we'll get:
$\displaystyle \sum_{n=1}^{N} (1-x)^2x^n =$

$\displaystyle \frac{(1-x)^2x(x^N-1)}{x-1}=$

$\displaystyle (x-1)x(x^N-1)=(x^N-x)(x-1)=x^{n+1}-x^N-x^2+x$ and then:[/tex]

$\displaystyle A(N,x)=\left| (\sum_{n=0}^N (1-x)^2 x^n) - \frac{x}{1-x} \right|=$

$\displaystyle \left| (x^{N+1}-x^N-x^2+x)(1-x)-x \right|=$

And I've no idea what should I do from here on....

I hope you'll be able to help me....

Thanks
• Jan 19th 2010, 08:08 AM
Drexel28
Quote:

Originally Posted by WannaBe
The series actually goes from 1 to infinity...
Hence:
$\displaystyle \sum_{n=1}^{\infty} (1-x)^2 x^n = \frac{x}{1-x}$
And:
$\displaystyle A(N,x)=\left| \left(\sum_{n=0}^N (1-x)^2 x^n\right) - \frac{x}{1-x} \right|$

If we simplify a little bit, we'll get:
$\displaystyle \sum_{n=1}^{N} (1-x)^2x^n =$

$\displaystyle \frac{(1-x)^2x(x^N-1)}{x-1}=$

$\displaystyle (x-1)x(x^N-1)=(x^N-x)(x-1)=x^{n+1}-x^N-x^2+x$ and then:[/tex]

$\displaystyle A(N,x)=\left| (\sum_{n=0}^N (1-x)^2 x^n) - \frac{x}{1-x} \right|=$

$\displaystyle \left| (x^{N+1}-x^N-x^2+x)(1-x)-x \right|=$

And I've no idea what should I do from here on....

I hope you'll be able to help me....

Thanks

CB showed you that $\displaystyle A(N,x)\leqslant (1-x)x^{N+1}\leqslant \left(1-\frac{N+1}{N+2}\right)\left(\frac{N+1}{N+2}\right) ^{N+1}\to0$. Note that [tex]\(N+1)x^{N+1}
• Jan 19th 2010, 09:02 AM
WannaBe
Yes, but:
Actually, I realy don't think that he showed this inequality considering the equality mark denoted in his msg...
Moreover, his calculation was :
$\displaystyle \sum_{n=0}^{\infty}$ and not $\displaystyle \sum_{n=1}^{\infty}$ and this is why I re-calculated everything and tried to figure out how to continue...
• Jan 19th 2010, 02:11 PM
CaptainBlack
Quote:

Originally Posted by WannaBe
Actually, I realy don't think that he showed this inequality considering the equality mark denoted in his msg...
Moreover, his calculation was :
$\displaystyle \sum_{n=0}^{\infty}$ and not $\displaystyle \sum_{n=1}^{\infty}$ and this is why I re-calculated everything and tried to figure out how to continue...

The convergence does not care about addition of a constant or a well behaved function. Proof of uniform convergence for a sum for 1 to infinity and form 0 to infinity are the same as long as that missing term is not too badly behaved.

Also you may be ought to consider being more careful to post the question you actually want answered. Or do you think our time worthless because you are not paying for it?

CB
• Jan 19th 2010, 11:38 PM
WannaBe
I realy don't want to answer my questions... I only want hints ...

Thanks a lot to both of you...